Codeforces Round #321 (Div. 2)-A. Kefa and First Steps，暴力水过~~

A. Kefa and First Steps

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Kefa decided to make some money doing business on the Internet for exactly n days. He knows that on the i-th
day (1 ≤ i ≤ n) he makes ai money.
Kefa loves progress, that's why he wants to know the length of the maximum non-decreasing subsegment in sequence ai.
Let us remind you that the subsegment of the sequence is its continuous fragment. A subsegment of numbers is called non-decreasing if all numbers in it follow in the non-decreasing order.

Help Kefa cope with this task!

Input

The first line contains integer n (1 ≤ n ≤ 105).

The second line contains n integers a1,  a2,  ...,  an (1 ≤ ai ≤ 109).

Output

Print a single integer — the length of the maximum non-decreasing subsegment of sequence a.

Examples

input

6
2 2 1 3 4 1

output

3

input

3
2 2 9

output

3

Note

In the first test the maximum non-decreasing subsegment is the numbers from the third to the fifth one.

In the second test the maximum non-decreasing subsegment is the numbers from the first to the third one.

这个题有点像求非单调递减子序列的长度，但仔细看看题发现是求连续一段非单减子数组的长度，这样和求单调递增子序列的方法很像，但看到数据范围两层循环遍历肯定会超时，题目要求的是子区间非单减的长度，为何要遍历呢？只需判断当前输入的值是否大于等于前一个的值，如果是，则此时的长度就等于前一个长度加一，反之，则此时长度为一，然后继续。看代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=100000+10;
int a[N],b[N];
int main()
{
    int n,i,j;
    while(~scanf("%d",&n))
    {
        memset(b,0,sizeof(a));
        int maxx=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            b[i]=1;
            for(j=i;j>=1;j--)
            {
                if(a[j]>=a[j-1]&&j!=1)
                    b[i]=b[j-1]+1;//用到了点动归思想；
                    break;//这样就不会超时了，甚至可以简化之；
            }
            maxx=max(maxx,b[i]);
        }
        printf("%d\n",maxx);
    }
    return 0;
}